Problem: Evaluate the triple integral. $ \int_2^4 \int_0^1 \int_{-2}^1 z^2 + x \, dx \, dy \, dz =$
Solution: We can evaluate triple integrals by repeated integration: $ \int_{a_0}^{a_1} \int_{b_0}^{b_1} \int_{c_0}^{c_1} f(x, y, z) \, dx \, dy \, dz = \int_{a_0}^{a_1} \left( \int_{b_0}^{b_1} \left[ \int_{c_0}^{c_1} f(x, y, z) \, dx \right] dy \right) dz$ The first layer: $\begin{aligned} &\int_2^4 \int_0^1 \int_{-2}^1 z^2 + x \, dx \, dy \, dz \\ \\ &= \int_2^4 \int_0^1 \left[ z^2x + \dfrac{x^2}{2} \right]_{-2}^1 dy \, dz \\ \\ &=\int_2^4 \int_0^1 \left( z^2 + \dfrac{1}{2} \right) - \left( -2z + 2 \right) dy \, dz \\ \\ &= \int_2^4 \int_0^1 3z^2 - \dfrac{3}{2} \, dy \, dz \end{aligned}$ The second layer: $\begin{aligned} \int_2^4 \int_0^1 3z^2 - \dfrac{3}{2} \, dy \, dz &= \int_2^4 \left[ 3z^2y - \dfrac{3}{2} \right]_0^1 dz \\ \\ &= \int_2^4 3z^2 - \dfrac{3}{2} \, dz \end{aligned}$ The third layer: $\begin{aligned} \int_2^4 3z^2 - \dfrac{3}{2} \, dz &= \left[ z^3 - \dfrac{3z}{2} \right]_2^4 \\ \\ &= (64 - 6) - (8 - 3) \\ \\ &= 53 \end{aligned}$ In conclusion: $ \int_2^4 \int_0^1 \int_{-2}^1 z^2 + x \, dx \, dy \, dz = 53$